Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
28 May 2020, 05:06
Kudos
Bookmarks
Hello guys,
I have a query regarding averages so I will try to ask the question by providing some examples to make myself clear.
Let's say that we have 2 sets A and B and we want to find the average of both those sets.
set A (1,3,8) set B (1,2,3)
Average of A=4
Average of B= 2
Average of both sets (4+2)/2=3
Alternatively (1+3+8+1+2+3)/6=3
As we can see in both cases the average is the same
However there are cases in which the above 2 approaches have 2 different results
set A ( 1,5,7,9) set B(13,14)
Average of set A= 5.5
Average of set B= 13.5
Average of set A and B= 19/2= 9.5
Alternatively (1+5+7+9+13+14)/6=8.16
Could someone shed some light on the issue that I have?
I have a query regarding averages so I will try to ask the question by providing some examples to make myself clear.
Let's say that we have 2 sets A and B and we want to find the average of both those sets.
set A (1,3,8) set B (1,2,3)
Average of A=4
Average of B= 2
Average of both sets (4+2)/2=3
Alternatively (1+3+8+1+2+3)/6=3
As we can see in both cases the average is the same
However there are cases in which the above 2 approaches have 2 different results
set A ( 1,5,7,9) set B(13,14)
Average of set A= 5.5
Average of set B= 13.5
Average of set A and B= 19/2= 9.5
Alternatively (1+5+7+9+13+14)/6=8.16
Could someone shed some light on the issue that I have?
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
Updated on: 15 Jul 2020, 10:28
Originally posted by BrushMyQuant on 29 May 2020, 03:49.
Last edited by BrushMyQuant on 15 Jul 2020, 10:28, edited 1 time in total.
Last edited by BrushMyQuant on 15 Jul 2020, 10:28, edited 1 time in total.
Kudos
Bookmarks
Average of two sets = Sum of all the numbers in both the sets/Total Number of numbers in both the sets
Example 1:
In this case you have the same number of elements in both the sets (3 each)
So when you are trying to do (Average of A + Average of B)/2 then it is still equal to average of both the sets because
((Average of A)*3 + (Average of B)*3) / 6 will give you the same result.
So, if you have same number of elements in two or more sets and if you want to take the average of all the sets together then you can still take the average of individual sets and then take average of all the sets to get average of all the sets together.
Example 2:
In this case you have different number of elements in both the sets (First has 4 and other has 2)
If you want to use the averages calculated for individual sets then you can do the following:
(4 *(Average of A) + 2*(Average of B)) / 6
But if you notice then 4*Average of A is same as Sum of all elements of A. SO, you can directly do some of all the elements in both the sets / total number of elements in both the sets.
When we find averages of two or more sets then you can also think about it as weighted average, instead of thinking of it as just average. Here weight is nothing but the total number of elements.
Hope it helps!
Example 1:
In this case you have the same number of elements in both the sets (3 each)
So when you are trying to do (Average of A + Average of B)/2 then it is still equal to average of both the sets because
((Average of A)*3 + (Average of B)*3) / 6 will give you the same result.
So, if you have same number of elements in two or more sets and if you want to take the average of all the sets together then you can still take the average of individual sets and then take average of all the sets to get average of all the sets together.
Example 2:
In this case you have different number of elements in both the sets (First has 4 and other has 2)
If you want to use the averages calculated for individual sets then you can do the following:
(4 *(Average of A) + 2*(Average of B)) / 6
But if you notice then 4*Average of A is same as Sum of all elements of A. SO, you can directly do some of all the elements in both the sets / total number of elements in both the sets.
When we find averages of two or more sets then you can also think about it as weighted average, instead of thinking of it as just average. Here weight is nothing but the total number of elements.
Hope it helps!
29 May 2020, 04:10
Kudos
Bookmarks
If you're combining two sets, and finding the average of the combined set, then you're always calculating what is known as a "weighted average". The average is "weighted" by the sizes of the sets (by the ratio of their sizes, technically). So if one set is twice as big as the other, the average you calculate will be "twice as close" to the bigger set's average -- the bigger set is twice as important as the smaller one.
So if you wanted to find the average income for people in the USA and in Canada, then since the USA has about ten times as many people as Canada, the USA's average is going to be far more important than Canada's average. You could not just find the USA average, and the Canada average, and then average those two numbers. You'd need to do a weighted average calculation. Weighted averages are too big a topic to get into in a short post, but one way to get the right answer is to give the USA average a weight of 10 (i.e. multiply it by 10) and the Canada average a weight of 1, and then divide by 10 + 1 = 11. So if average income in the US is $31,000 and in Canada is $20,000, then the overall average will be
[(10)(31,000) + (1)(20,000)] / 11 = 30,000
If you look at how far the answer here is from the individual averages, you'll notice it's 10 times as far from Canada's average than from the USA average. In non-grammatical speak, you could say the answer is "10 times as close" to the USA average (that's not correct English though). That's identical to the ratio of the sizes of the two populations, and weighted averages always work out that way. That observation is the basis for a different weighted average method that is sometimes called "alligation", which you could look into, though that method takes a bit of practice. If you can learn it though, it's the best weighted average method for GMAT purposes.
So if you wanted to find the average income for people in the USA and in Canada, then since the USA has about ten times as many people as Canada, the USA's average is going to be far more important than Canada's average. You could not just find the USA average, and the Canada average, and then average those two numbers. You'd need to do a weighted average calculation. Weighted averages are too big a topic to get into in a short post, but one way to get the right answer is to give the USA average a weight of 10 (i.e. multiply it by 10) and the Canada average a weight of 1, and then divide by 10 + 1 = 11. So if average income in the US is $31,000 and in Canada is $20,000, then the overall average will be
[(10)(31,000) + (1)(20,000)] / 11 = 30,000
If you look at how far the answer here is from the individual averages, you'll notice it's 10 times as far from Canada's average than from the USA average. In non-grammatical speak, you could say the answer is "10 times as close" to the USA average (that's not correct English though). That's identical to the ratio of the sizes of the two populations, and weighted averages always work out that way. That observation is the basis for a different weighted average method that is sometimes called "alligation", which you could look into, though that method takes a bit of practice. If you can learn it though, it's the best weighted average method for GMAT purposes.
